# Prove complements of independent events are independent.

Given a finite set of events $\{A_i\}$ which are mutually independent, i.e., for every subset $\{A_n\}$, $$\mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i).$$

show that the set $\{A_i^c\}$, that is the set of complements of the original events, is also mutually independent.

I can prove this, but my proof relies on the Inclusion-Exclusion principle (as does the proof given in this question). I'm hoping there is a more concise proof.

Can this statement be proved without the use of the Inclusion-Exclusion principle?

Hint: prove that the set of events stays independent if you replace one of them by its complement, i.e. that given your conditions the set $\{A_1^c, A_2, \ldots, A_n\}$ is independent. Then use this $n$ times to replace all of $A_i$ by their complements one by one.

Update. Hint 2: to avoid clutter, let me show you what I mean on the example of two events, $B$ and $C$. Suppose $B$ and $C$ are independent, i.e. $P(B \cap C) = P(B) \cdot P(C)$. We want to show that $B^c$ and $C$ are independent. Indeed: \begin{align*} P(B^c \cap C) &= P(C \setminus (B \cap C)) \\ &= P(C) - P(B \cap C) \\ &= P(C) - P(B) \cdot P(C) \\ &= P(C) \cdot (1-P(B)) \\ &= P(B^c) \cdot P(C). \end{align*} I didn't use inclusion-exclusion here. And this approach scales, i.e. it works the same if you consider more than $2$ variables.

• Thanks for the hint, I hadn't thought of doing it this way. But unfortunately I'm still stuck - I can't see how to show $\{A_1^c, A_2, \ldots, A_n\}$ is independent without using the I-E principle. – Doria Mar 17 '15 at 0:57
• Great hint Dan! Love this version of the proof. – Mehness Apr 26 '18 at 20:44
• Can this argument be used for a contable family of independent events? – RLC Feb 3 at 16:18

Let $[n] = \{1, \cdots, n\}$ and $p(X_1, \cdots, X_n) = \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} X_i$. That is, $p$ is a polynomial in $X_1, \cdots, X_n$ consisting of square-free monomials. If $\{A_1, \cdots, A_n\}$ are mutually independent, then

\begin{align*} \Bbb{E}[p(\mathbf{1}_{A_1}, \cdots, \mathbf{1}_{A_n})] &= \sum_{I \subseteq [n] } a_{I} \Bbb{E}\Big( \prod_{i \in I} \mathbf{1}_{A_i} \Big) \\ &= \sum_{I \subseteq [n] } a_{I} \Bbb{P}\Big( \bigcap_{i \in I} {A_i} \Big) \\ &= \sum_{I \subseteq [n] } a_{I} \prod_{i \in I} \Bbb{P} (A_i) \\ &= p(\Bbb{P}(A_1), \cdots, \Bbb{P}(A_n)), \end{align*}

Then the conclusion follows for the choice $p(X_1, \cdots, X_n) = (1-X_1)\cdots(1-X_n)$.